Question

$\int \frac{(x+3)e^x}{(x+4{)}^2}dx$ is equal to

• A. $\frac{1}{(x+4{)}^2}+C$
• B. $\frac{e^x}{(x+3)}+C$
• C. $\frac{e^x}{(x+4{)}^2}+C$
• D. $\frac{e^x}{(x+4)}$

Answer 1

The correct option is D $\frac{e^x}{(x+4)}$

$\int \frac{(x+3)}{{(x+4)}^2}{e}^{x}dx$
$=\int e^x\frac{(x+3+1-1)}{{(x+4)}^2}dx$
$=\int e^x[\frac{(x+4)}{{(x+4)}^2}-\frac{1}{{(x+4)}^2}]dx$
$=\int e^x[\frac{1}{x+4}-\frac{1}{{(x+4)}^2}]dx$
$=\int e^x[\frac{1}{x+4}+\left[\frac{-1}{{(x+4)}^2}\right]]dx$
Now, $\int e^x(f\left(x\right)+f^{{}^{\prime }}\left(x\right))dx=e^{x}f\left(x\right)$
$\therefore I=\frac{e^x}{(x+4)}$